When the spring is stretched by 15.2 cm = 0.152 m, the spring exerts a restorative force with magnitude (due to Hooke's law)
where is the spring constant. Solve for
.
The amount of work required to stretch or compress a spring by from equilibrium length is
Then the work needed to stretch the spring by 15.2 cm is
and by 15.2 + 13.7 = 28.9 cm is
so the work needed to stretch from 15.2 cm to 28.9 cm from equilibrium is
Answer:
456.4 N
Explanation:
From conservation of energy, the potential energy is converted to kinetic energy hence
mgh=½mv²
Making v the subject then
Where g is acceleration due to gravity and h is height whike v is final velocity. Substituting 4.9m for h and 9.81 m/s² for g then
Change in momentum equals to the impulse.
Impulse, I= Ft
Change in momentum, ∆p= m(v-u)
Ft=m(v-u) making F the subject of formula then
Where F is force in Newton, t is time in seconds, m is mass of diver, v and u are the final and initial velocities respectively.
Substituting 68 kg for m, 9.8 m/s for v, 0 m/s for u since it is initially at rest and 1.46 s for t
Answer:
(1) 10^−2 m
Explanation:
The diameter of the tire of an automobile is generally expressed in centimetres; we can say that the diameter of a tire is generally about
d = 20 cm (20 centimetres)
Now we have to verify which option is closest to this value. To do that, we have to keep in mind the equivalence between metres and centimetres; in fact, we have:
This means that we can rewrite the diameter of the tire of a car as
By comparing it with the given options, we see that the closest option is
(1) 10^−2 m
which is therefore the correct answer.